Inequality about Eigenvalue of a Real Symmetric Matrix

Problem 451

Let $A$ be an $n\times n$ real symmetric matrix.
Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.\]

Proof.

Since these eigenvalues are real numbers, there is the largest one.
Let $\lambda$ be the largest eigenvalue of $A$.
With this choice of $\lambda$ we show that the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2\]
holds for any $\mathbf{v}\in \R^n$.

Also recall that for a real symmetric matrix, there are eigenvalues $\mathbf{v}_1, \dots, \mathbf{v}_n$ corresponding to $\lambda_1, \dots, \lambda_n$ such that
\[B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}\]
form an orthonormal basis of $\R^n$.
(This statement is equivalent to that every real symmetric matrix is diagonalizable by an orthogonal matrix.)

Let $\mathbf{v}$ be an arbitrary vector in $\R^n$.
Then since $B$ is a basis of $\R^n$, we can write
\[\mathbf{v}=c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n\]
for some $c_1, \dots, c_n\in \R$.

Since $\lambda$ is the largest eigenvalue of $A$, we have
\begin{align*}
\mathbf{v}\cdot A\mathbf{v}&=c_1^2\lambda_1+\cdots+c_n^2\lambda_n\\
& \leq c_1^2\lambda+\cdots+c_n^2\lambda\\
&=\lambda(c_2^2+\cdots+c_n^2)\\
&=\lambda \|\mathbf{v}\|^2.
\end{align*}
Hence the required inequality holds.

Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix
Let $A$ be an $n\times n$ matrix with real number entries.
Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.
Proof.
Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$.
The orthogonality of the […]

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